\newproblem{lay:5_4_23}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.4.23}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	If $B=P^{-1}AP$ and $\mathbf{x}$ is an eigenvector of $A$ corresponding to an eigenvalue $\lambda$, then $P^{-1}\mathbf{x}$ is an eigenvector of $B$
	corresponding also to an eigenvalue $\lambda$.
}{
  % Solution
	Let's check whether the statement proposed by the problem is true or not. If it is true, it means that
	\begin{center}
		$B(P^{-1}\mathbf{x})=\lambda(P^{-1}\mathbf{x})$
	\end{center}
	According to the problem we have that $B=P^{-1}AP$, so
	\begin{center}
		$B(P^{-1}\mathbf{x})=(P^{-1}AP)(P^{-1}\mathbf{x})=P^{-1}A\mathbf{x}$
	\end{center}
	But by hypothesis $\mathbf{x}$ is an eigenvector of $A$ corresponding to an eigenvalue $\lambda$, that is $A\mathbf{x}=\lambda\mathbf{x}$. Consequently,
	\begin{center}
		$P^{-1}A\mathbf{x}=P^{-1}(\lambda\mathbf{x})=\lambda(P^{-1}\mathbf{x})$
	\end{center}
	Finally, we have proven that, as stated by the problem,
	\begin{center}
		$B(P^{-1}\mathbf{x})=\lambda(P^{-1}\mathbf{x})$
	\end{center}
}
\useproblem{lay:5_4_23}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
